# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 5 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Jul 4, 2018

The distance is $= 1.349 m$

#### Explanation:

The equation of the trajectory of the projectile in the vertical plane, $O$ being the origin is

$y = x \tan \theta - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \theta}\right) {x}^{2}$

The initial velocity is $u = 5 m {s}^{-} 1$

The angle of projection is $\theta = \frac{5}{12} \pi$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Therefore,

$y = x \tan \left(\frac{5}{12} \pi\right) - \left(\frac{9.8}{2 \cdot {5}^{2} \cdot {\cos}^{2} \left(\frac{5}{12} \pi\right)}\right) {x}^{2}$

$y = 3.732 x - 2.926 {x}^{2}$

The highest point is reached when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3.732 - 5.852 x$

$3.732 - 5.852 x = 0$

$\implies$, $x = \frac{3.732}{5.852} = 0.638$

And

$y = 3.732 \cdot 0.638 - 2.926 \cdot {0.638}^{2} = 1.189$

The highest point is $= \left(0.638 , 1.189\right)$

The distance from the stat¡rting point is

$d = \sqrt{{0.638}^{2} + {1.189}^{2}} = 1.349 m$

graph{(3.732x-2.926x^2) [-0.988, 2.43, -0.128, 1.581]}